1130 POJ#1376 Robot

摘要

-模拟机器人行走，找到耗时最短路径

原题目摘要

-POJ1376 Robot http://poj.org/problem?id=1376

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The Robot Moving Institute is using a robot in their local store to transport different items. Of course the robot should spend only the minimum time necessary when travelling from one place in the store to another. The robot can move only along a straight line (track). All tracks form a rectangular grid. Neighbouring tracks are one meter apart. The store is a rectangle N x M meters and it is entirely covered by this grid. The distance of the track closest to the side of the store is exactly one meter. The robot has a circular shape with diameter equal to 1.6 meter. The track goes through the center of the robot. The robot always faces north, south, west or east. The tracks are in the south-north and in the west-east directions. The robot can move only in the direction it faces. The direction in which it faces can be changed at each track crossing. Initially the robot stands at a track crossing. The obstacles in the store are formed from pieces occupying 1m x 1m on the ground. Each obstacle is within a 1 x 1 square formed by the tracks. The movement of the robot is controlled by two commands. These commands are GO and TURN. 
The GO command has one integer parameter n in {1,2,3}. After receiving this command the robot moves n meters in the direction it faces. 

The TURN command has one parameter which is either left or right. After receiving this command the robot changes its orientation by 90o in the direction indicated by the parameter. 

The execution of each command lasts one second. 

Help researchers of RMI to write a program which will determine the minimal time in which the robot can move from a given starting point to a given destination. 

Input

The input consists of blocks of lines. The first line of each block contains two integers M <= 50 and N <= 50 separated by one space. In each of the next M lines there are N numbers one or zero separated by one space. One represents obstacles and zero represents empty squares. (The tracks are between the squares.) The block is terminated by a line containing four positive integers B1 B2 E1 E2 each followed by one space and the word indicating the orientation of the robot at the starting point. B1, B2 are the coordinates of the square in the north-west corner of which the robot is placed (starting point). E1, E2 are the coordinates of square to the north-west corner of which the robot should move (destination point). The orientation of the robot when it has reached the destination point is not prescribed. We use (row, column)-type coordinates, i.e. the coordinates of the upper left (the most north-west) square in the store are 0,0 and the lower right (the most south-east) square are M - 1, N - 1. The orientation is given by the words north or west or south or east. The last block contains only one line with N = 0 and M = 0.

Output

The output contains one line for each block except the last block in the input. The lines are in the order corresponding to the blocks in the input. The line contains minimal number of seconds in which the robot can reach the destination point from the starting point. If there does not exist any path from the starting point to the destination point the line will contain -1.

 

Sample Input

9 100 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 1 00 0 0 1 0 0 0 0 0 00 0 1 0 0 0 0 0 0 00 0 0 0 0 0 1 0 0 00 0 0 0 0 1 0 0 0 00 0 0 1 1 0 0 0 0 00 0 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 1 07 2 2 7 south0 0

Sample Output

12

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题目理解

-机器人 占用两个格，行走需要考虑能否前行；寻找通过一定步骤的行走到达目标坐标的路线；使用bfs进行最短路的寻找；

注意

去过的状态不要重复

旋转方向的控制量范围保持好

机器人行走中的道路是否可行检测

日期

-2017 11 30

附加

-

代码

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#include <algorithm>#include <cstdio>#include <iostream>#include <stack>#include <queue>#include <memory>#include <list>#define MAX 50using namespace std;int m,n;int b1,b2,e1,e2,d0;int dx[]={-1,0,1,0};int dy[]={0,1,0,-1};struct Status{    bool dir[4];    void init(){dir[0]=dir[1]=dir[2]=dir[3]=false;}};struct QStatus{// single state store in queue    int _x, _y;    int d;//direction    int step;};bool Matrix[MAX][MAX];Status Vis_Matrix[MAX][MAX];bool check_point_border(int x,int y){    return x>=0&&x<m        && y>=0&&y<n;}bool vis_check_path(int x,int y,int d){//可以前行    return check_point_border(x-1,y-1)&& !Matrix[x-1][y-1]        && check_point_border(x-1,y)&& !Matrix[x-1][y ]        && check_point_border(x,y-1)&& !Matrix[x ][y-1]        && check_point_border(x,y)&& !Matrix[x ][y ];}bool Go(QStatus qs,int length){    if(vis_check_path(qs._x+length*dx[qs.d],qs._y+length*dy[qs.d],qs.d)) return true;    else return false;}bool Turn(QStatus &qs,int ro){    qs.d=(qs.d+ro+4)%4;    return !Vis_Matrix[qs._x][qs._y].dir[qs.d] ;}void addStatus(queue<QStatus> &q,QStatus qs){    qs.d = (qs.d+4)%4;    if(!Vis_Matrix[qs._x][qs._y].dir[qs.d]){        Vis_Matrix[qs._x][qs._y].dir[qs.d] = true;        q.push(qs);//cout<<"\tADD:"<<qs._x<<" "<<qs._y<<" dir:"<<qs.d<<" step:"<<qs.step<<endl;    }}int find_min_path(){    queue<QStatus> Q;    addStatus(Q,{b1,b2,d0,0});    while(!Q.empty()){        QStatus &qs = Q.front();//cout<<qs._x<<" "<<qs._y<<" dir:"<<qs.d<<" step:"<<qs.step<<endl;        if(qs._x==e1&&qs._y==e2){            return qs.step;        }        for(int i=1;i<=3;i++){             if(Go(qs,i)){//只要可行 就要测试 2,3                addStatus(Q,{qs._x+i*dx[qs.d],qs._y+i*dy[qs.d],qs.d,qs.step+1});             }else break;        }        addStatus(Q,{qs._x,qs._y,qs.d-1,qs.step+1});        addStatus(Q,{qs._x,qs._y,qs.d+1,qs.step+1});        Q.pop();    }    return -1;}void solve(){    int tmp;    for(int i=0;i<m;i++){        for(int j=0;j<n;j++){            scanf("%d",&tmp);            Matrix[i][j]=tmp;            Vis_Matrix[i][j].init();        }    }    scanf("%d%d%d%d",&b1,&b2,&e1,&e2);    char dir[5];    scanf("%s",dir);    switch(dir[0]){        case 'n':d0=0;break;        case 'e':d0=1;break;        case 's':d0=2;break;        case 'w':d0=3;break;    }    if(b1==e1&&b2==e2) printf("0\n");    else{        int ans = find_min_path();        if(ans==0) printf("-1\n");        else printf("%d\n",ans);    }}int main(){       while(scanf("%d%d",&m,&n)){        if(m>0 && n>0) solve();        else break;    }    return 0;}